# Is $\int_a^a f(x) dx$ always zero?

Is the result:

$$\int_a^a f(x) \,\text{d}x = 0$$

always zero?

This seems obvious at first, but what if $f(x)$ diverges at $x=a$?

For example, Wolfram Alpha tells me

$$\int_0^0 \frac{1}{x}\,\text{d}x = 0\qquad (1)$$

But: $$\int_0^{1\times10^{-40}}\frac{1}{x}\,\text{d}x = \infty\qquad (2)$$

As far as I can tell, this happens no matter how small I set the upper limit of integration.

Is $(1)$ correct? Why is it true, even if both $f$ and it's antiderivative diverge at the point?

If it's not true in general, what conditions must $f(x)$ satisfy?

Edit

I do not believe my question is a true duplicate of the other. While the question about integrating on a point is shared, my question is more general, since that one concentrates on s specific (complicated) integral. I ask what is needed for general $f(x)$. If you are the answers to that question, neither is generalizable to other integrands.

• I suggest that you ask the main question as: What conditions should $f(x)$ satisfy in order to have $\int_a^a f(x) dx = 0$. – user230734 Jul 15 '15 at 2:18
• What if you had a function that was defined only on $x = a$ and no where else? What if it is not only defined on $x = a$, but it is infinite at that point? hitoshi.berkeley.edu/221a/delta.pdf – Carlos - the Mongoose - Danger Jul 15 '15 at 2:19
• The usual definition for Lebesgue integration, which is a wider class of integration than Riemann Integration is that $0\cdot\infty=0$. So (1) and (2) are correct. – Alex R. Jul 15 '15 at 2:22
• Good question. Briefly, the Riemann integral make sense only when the function $f$ is bounded over the domain of integration. So, I don't think (1) makes sense as a Riemann integral. As a generalized integral, you cannot approach (1) by integrals like (2), since (2) is itself an improper integral. The problem really is the lower limit of integration: the function $1/x$ is not defined at zero. However, in measure theory, $\int_a^a f d\mu=0$ is always correct. – Joe Aaron Jul 15 '15 at 2:28
• @JoeAaron: Techincally, you still have to define $f(x)=1/x$ to be $\textit something$ at $x=0$ for $\int_a^a f d\mu=0$ to make sense. – Matematleta Jul 15 '15 at 2:36

Let $M$ be a subset of $\mathbb R$ such that $M$ has Lebesgue measure $0$. Let $f$ be a function on $M$ such that $f(x) = \infty$ for all $x\in M$. I claim $\int_M f(x) dx = 0$. The proof is as follows.
For an integer $n\ge 1$, let $f_n$ be a constant $n$ function on $M$. Then $f_1\le f_2\le \cdots$ and $\text{lim}_{n\rightarrow \infty} f_n(x) = f(x)$ for all $x\in M$. By Lebesgue monotone convergence theorem, $\int_M f(x) dx = \text{lim}_{n\rightarrow \infty} \int_M f_n(x) dx$. The right-hand side of the equation is obviously $0$.
The example $f(x) = 1/x$ is not (Lebesgue) integrable on $(0, a)$ for any $a>0$, so $a\mapsto\int_0^a f(x) \mathrm d x$ need not to be continuous at $a=0$.
However, if $f$ is integrable, then $a\mapsto\int_0^a f(x) \mathrm d x$ is continuous in $a$ and $$\lim_{a\to 0} \int_0^a f(x) \mathrm d x = \int_0^0 f(x) \mathrm d x = 0.$$